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Sunday, February 22, 2009

M Theory Lesson 264

Recall that circulants are always magic as well as square magic in the sense that the sum of squares along a row or column is a fixed constant. In particular, any Koide mass matrix $M$ has this property.

But MUB operators such as $F_3$ are not magic. For example, the action of $F_3 F_2$ (the neutrino mixing matrix) on $M$ results in a $1 \times 2$ block matrix in terms of the square roots of the masses, because $F_3$ diagonalises, and $F_2$ then acts on a pair of mass eigenvalues. The fact that this matrix is not magic is the same as the statement that $m_1 \neq m_2$. The fact that it is not square magic follows from the statement that $s < 0$ in

$\textrm{cos} \delta = \frac{s - 6v}{s}$,

where $\delta$ is the angle shared by all three masses. This property is shared by the hadron fits.

Yeah, sorry, Carl. It was too much trouble to write out the matrices today. Anyway, I'm not sure these kind of products will help. I started with the magic conditions on a certain matrix, and derived the conditions on s and v from there.